Add solution 270

Signed-off-by: Thomas Köhler <jean-luc@picard.franken.de>

4 files changed, 241 insertions, 0 deletions

diff --git a/challenge-270/jeanluc2020/blog-1.txt b/challenge-270/jeanluc2020/blog-1.txt
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+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-270-1.html
diff --git a/challenge-270/jeanluc2020/blog-2.txt b/challenge-270/jeanluc2020/blog-2.txt
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+http://gott-gehabt.de/800_wer_wir_sind/thomas/Homepage/Computer/perl/theweeklychallenge-270-2.html
diff --git a/challenge-270/jeanluc2020/perl/ch-1.pl b/challenge-270/jeanluc2020/perl/ch-1.pl
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+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-270/#TASK1
+#
+# Task 1: Special Positions
+# =========================
+#
+# You are given a m x n binary matrix.
+#
+# Write a script to return the number of special positions in the given binary matrix.
+#
+## A position (i, j) is called special if $matrix[i][j] == 1 and all other elements in the row i and column j are 0.
+#
+## Example 1
+##
+## Input: $matrix = [ [1, 0, 0],
+##                    [0, 0, 1],
+##                    [1, 0, 0],
+## ]
+## Output: 1
+##
+## There is only one special position (1, 2) as $matrix[1][2] == 1
+## and all other elements in row 1 and column 2 are 0.
+#
+## Example 2
+##
+## Input: $matrix = [ [1, 0, 0],
+##                    [0, 1, 0],
+##                    [0, 0, 1],
+## ]
+## Output: 3
+##
+## Special positions are (0,0), (1, 1) and (2,2).
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# We walk the matrix and for each position we check whether
+# the position is special.
+# For this we use a function that just checks the same column
+# in all rows, and checks the same row for all columns. If
+# row and column are the target row and column, we conclude the
+# point is not special if it isn't 1, for all other values we
+# conclude the point is not special if it isn't 0. If we reach
+# the end and we didn't eliminate the point as not special, then
+# we conclude this point is special.
+
+use strict;
+use warnings;
+
+special_positions( [ [1, 0, 0], [0, 0, 1], [1, 0, 0], ] );
+special_positions( [ [1, 0, 0], [0, 1, 0], [0, 0, 1], ] );
+
+sub special_positions {
+   my $matrix = shift;
+   my @rows = @$matrix;
+   my $num_rows = scalar(@rows);
+   my $num_columns = scalar(@{$rows[0]});
+   my $special = 0;
+   foreach my $i (0..$num_rows-1) {
+      foreach my $j (0..$num_columns-1) {
+         if(is_special($i, $j, $num_rows, $num_columns, $matrix)) {
+            $special++;
+         }
+      }
+   }
+   print "Output: $special\n";
+}
+
+sub is_special {
+   my ($i, $j, $num_rows, $num_columns, $matrix) = @_;
+   foreach my $row (0..$num_rows-1) {
+      if($i == $row) {
+         return 0 if $matrix->[$row]->[$j] != 1;
+      } else {
+         return 0 if $matrix->[$row]->[$j] != 0;
+      }
+   }
+   foreach my $column (0..$num_columns-1) {
+      if($j == $column) {
+         return 0 if $matrix->[$i]->[$column] != 1;
+      } else {
+         return 0 if $matrix->[$i]->[$column] != 0;
+      }
+   }
+   return 1;
+}
+
diff --git a/challenge-270/jeanluc2020/perl/ch-2.pl b/challenge-270/jeanluc2020/perl/ch-2.pl
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index 0000000000..8356ab0209
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+#!/usr/bin/perl
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-270/#TASK2
+#
+# Task 2: Distribute Elements
+# ===========================
+#
+# You are give an array of integers, @ints and two integers, $x and $y.
+#
+# Write a script to execute one of the two options:
+#
+## Level 1:
+## Pick an index i of the given array and do $ints[i] += 1
+##
+## Level 2:
+## Pick two different indices i,j and do $ints[i] +=1 and $ints[j] += 1.
+#
+# You are allowed to perform as many levels as you want to make every elements
+# in the given array equal. There is cost attach for each level, for Level 1,
+# the cost is $x and $y for Level 2.
+#
+# In the end return the minimum cost to get the work done.
+#
+## Example 1
+##
+## Input: @ints = (4, 1), $x = 3 and $y = 2
+## Output: 9
+##
+## Level 1: i=1, so $ints[1] += 1.
+## @ints = (4, 2)
+##
+## Level 1: i=1, so $ints[1] += 1.
+## @ints = (4, 3)
+##
+## Level 1: i=1, so $ints[1] += 1.
+## @ints = (4, 4)
+##
+## We perforned operation Level 1, 3 times.
+## So the total cost would be 3 x $x => 3 x 3 => 9
+#
+## Example 2
+##
+## Input: @ints = (2, 3, 3, 3, 5), $x = 2 and $y = 1
+## Output: 6
+##
+## Level 2: i=0, j=1, so $ints[0] += 1 and $ints[1] += 1
+## @ints = (3, 4, 3, 3, 5)
+##
+## Level 2: i=0, j=2, so $ints[0] += 1 and $ints[2] += 1
+## @ints = (4, 4, 4, 3, 5)
+##
+## Level 2: i=0, j=3, so $ints[0] += 1 and $ints[3] += 1
+## @ints = (5, 4, 4, 4, 5)
+##
+## Level 2: i=1, j=2, so $ints[1] += 1 and $ints[2] += 1
+## @ints = (5, 5, 5, 4, 5)
+##
+## Level 1: i=3, so $ints[3] += 1
+## @ints = (5, 5, 5, 5, 5)
+##
+## We perforned operation Level 1, 1 time and Level 2, 4 times.
+## So the total cost would be (1 x $x) + (3 x $y) => (1 x 2) + (4 x 1) => 6
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# We find the index of the biggest number in the array, and the
+# indices of the two smallest ones. As long as the smallest number
+# is still smaller than the biggest one, we check:
+# - if the second smallest number is still smaller than the biggest
+#   AND the cost of incrementing two numbers is less than twice the
+#   cost of incrementing one number, increment the two smallest
+#   numbers
+# - else increment the smallest number
+#
+# Two helper functions are used: max_at() returns the index of the
+# biggest number in the array, min_at() returns the indices of the
+# two smallest numbers in the array (first one wins if two numbers have
+# the same value)
+
+use strict;
+use warnings;
+
+distribute_elements( [4, 1], 3, 2);
+distribute_elements( [2, 3, 3, 3, 5], 2, 1);
+
+sub distribute_elements {
+   my ($array, $x, $y) = @_;
+   my @ints = @$array;
+   print "Input: (", join(", ", @ints), "), \$x = $x, \$y = $y\n";
+   my $result = 0;
+   my $max_pos = max_at(@ints);
+   my ($min1_pos, $min2_pos) = min_at(@ints);
+   while($ints[$min1_pos] < $ints[$max_pos]) {
+      if($ints[$min2_pos] < $ints[$max_pos] && 2*$x > $y) {
+         $ints[$min2_pos]++;
+         $ints[$min1_pos]++;
+         $result += $y;
+      } else {
+         $ints[$min1_pos]++;
+         $result += $x;
+      }
+      ($min1_pos, $min2_pos) = min_at(@ints);
+   }
+   print "Output: $result\n";
+}
+
+sub max_at {
+   my @ints = @_;
+   my $max_pos = 0;
+   my $max = $ints[0];
+   foreach my $i (1..$#ints) {
+      if($ints[$i] > $max) {
+         $max = $ints[$i];
+         $max_pos = $i;
+      }
+   }
+   return $max_pos;
+}
+
+sub min_at {
+   my @ints = @_;
+   my ($min1_pos, $min2_pos, $min1, $min2);
+   if($ints[0] > $ints[1]) {
+      $min1_pos = 1;
+      $min2_pos = 0;
+      $min1 = $ints[1];
+      $min2 = $ints[0];
+   } else {
+      $min1_pos = 0;
+      $min2_pos = 1;
+      $min1 = $ints[0];
+      $min2 = $ints[1];
+   }
+   foreach my $i (2..$#ints) {
+      next unless $ints[$i] < $min2;
+      if($ints[$i] < $min1) {
+         ($min1, $min2) = ($ints[$i], $min1);
+         ($min1_pos, $min2_pos) = ($i, $min1_pos);
+      } else {
+         $min2 = $ints[$i];
+         $min2_pos = $i;
+      }
+   }
+   return ($min1_pos, $min2_pos);
+}
+